# Derivative and Definite Integral question

• Apr 8th 2006, 09:18 PM
nirva
Derivative and Definite Integral question
Derivative:
$\displaystyle p(t) = t^-^3 - 3t\sqrt{t^3}$
Is the square root of t^3 the same as t^1.5?

Integrals(Express the following as limits, don't evaluate):
As 0 goes to 4
$\displaystyle \int {\( (2x+1)^1^/^3} dx =$

As 4 goes to 7
$\displaystyle \int {\(\sqrt{2t} dt =$
• Apr 8th 2006, 10:57 PM
Bahama.Llama
Quote:

Originally Posted by nirva
Derivative:
$\displaystyle p(t) = t^-^3 - 3t\sqrt{t^3}$
Is the square root of t^3 the same as t^1.5?

square root t^3 is pretty much just t^3*.5 (when a power is raised to a power, you multiply the powers). So yes. If nobody has helped you by morning, I'll do what I can. Until then, nap time!
• Apr 9th 2006, 06:38 AM
ThePerfectHacker
Quote:

Originally Posted by nirva
Integrals(Express the following as limits, don't evaluate):
As 0 goes to 4
$\displaystyle \int {\( (2x+1)^1^/^3} dx =$

Use substitution rule. Let $\displaystyle u=2x+1$ then, the integral becomes,
$\displaystyle \frac{1}{2} \int^9_1 u^3 du$ the reason for the limits is because when $\displaystyle x=0$ then $\displaystyle u=1$ and when $\displaystyle x=4$ then $\displaystyle u=9$
Quote:

Originally Posted by Nirva
As 4 goes to 7
$\displaystyle \int {\(\sqrt{2t} dt =$

Use subsitution $\displaystyle x=2t$ then using the rule properly you have,
$\displaystyle \frac{1}{2}\int^{14}_{8}\sqrt{x}dx$